Jacobi elliptic functions

Remarks

Let $K = K(k)$ and $K' = K(\sqrt{1 - k^2})$ be complete elliptic integrals of the first kind. The extended functions are elliptic functions with these periods: $\operatorname{sn}$ has periods $4K$ and $2iK'$; $\operatorname{cn}$ has periods $4K$ and $2K + 2iK'$; $\operatorname{dn}$ has periods $2K$ and $4iK'$. They satisfy $$\operatorname{sn}^2 u + \operatorname{cn}^2 u = 1, \qquad \operatorname{dn}^2 u + k^2\operatorname{sn}^2 u = 1,$$ $$\frac{d}{du}\operatorname{sn} u = \operatorname{cn} u \operatorname{dn} u, \quad \frac{d}{du}\operatorname{cn} u = -\operatorname{sn} u \operatorname{dn} u, \quad \frac{d}{du}\operatorname{dn} u = -k^2\operatorname{sn} u \operatorname{cn} u.$$ Example: at $u = K$ one has $\operatorname{am} K = \pi/2$, so $\operatorname{sn} K = 1$, $\operatorname{cn} K = 0$ and $\operatorname{dn} K = k'$, where $k' = \sqrt{1 - k^2}$. Check of the first derivative rule: $\frac{d}{du}\sin\operatorname{am} u = \cos(\operatorname{am} u)\cdot\frac{d\operatorname{am}}{du} = \operatorname{cn} u \operatorname{dn} u$. Non-example: $\sin$ is not one of these functions for $0 < k < 1$, since $\sin$ is entire while $\operatorname{sn}$ has poles, at $u = iK'$ for example. Limits: for $k \to 0$ they become $\sin$, $\cos$, $1$, and the formulas with $k = 1$ give $\tanh$, $\operatorname{sech}$, $\operatorname{sech}$.